Page:On Governors.pdf/11

Let the velocity of a particle whose mass is $\scriptstyle \ m$ resolved in the direction of $\scriptstyle \ x$ be

(50)	${\frac {dx}{dt}}=p_{1}{\frac {d\xi }{dt}}+q_{1}{\frac {d\eta }{dt}}$

with similar expressions for the other coordinate directions, putting suffixes 2 and 3 to denote the values of $\scriptstyle \ p$ and $\scriptstyle \ q$ for these directions. Then Lagrange's equation of motion becomes

(51)	$\Xi \delta \xi +\mathrm {H} \delta \eta =\sum \left({\frac {d^{2}x}{dt^{2}}}\delta x+{\frac {d^{2}y}{dt^{2}}}\delta y+{\frac {d^{2}z}{dt^{2}}}\delta z\right)=0$

where $\scriptstyle \ \Xi$ and $\scriptstyle \ H$ are the forces tending to increase $\scriptstyle \ \xi$ and $\scriptstyle \ \eta$ respectively, no force being supposed to be applied at any other point.

Now putting

(52)	$\delta x=p_{1}d\xi +q_{1}d\eta$

and

(53)	${\frac {d^{2}x}{dt^{2}}}=p_{1}{\frac {d^{2}\xi }{dt^{2}}}+q_{1}{\frac {d^{2}\eta }{dt^{2}}}$

the equation becomes

(54)	$\left(\Xi -\sum mp^{2}{\frac {d^{2}\xi }{dt^{2}}}-\sum mpq{\frac {d^{2}\eta }{dt^{2}}}\right)\delta \xi +\left(H-\sum mpq{\frac {d^{2}\xi }{dt^{2}}}-\sum mq^{2}{\frac {d^{2}\eta }{dt^{2}}}\right)\delta \eta =0$

and since $\scriptstyle \ \delta x$ and $\scriptstyle \ \delta \eta$ are independent, the coefficient of each must be zero.

If we now put

(55)	$\sum (mp^{2})=L,\sum (mpq)=M,\sum (mpq^{2})=N$

where $\scriptstyle \ p^{2}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2},\ pq=p_{1}q_{1}+p_{2}q_{2}+p_{3}q_{3}$, and $\scriptstyle \ q^{2}=q_{1}^{2}+q_{2}^{2}+q_{3}^{2}$, the equations of motion will be

(56)	$\Xi =L{\frac {d^{2}\xi }{dt^{2}}}+M{\frac {d^{2}\eta }{dt^{2}}}$

(57)	$H=M{\frac {d^{2}\xi }{dt^{2}}}+N{\frac {d^{2}\eta }{dt^{2}}}$

If the apparatus is so arranged that $\scriptstyle \ M=0$, then the two motions will be independent of each other; and the motions indicated by $\scriptstyle \ \xi$ and $\scriptstyle \ \eta$ will be about conjugate axes—that is, about axes such that the rotation round one of them does not tend to produce a force about the other.

Now let $\scriptstyle \ \Theta$ be the driving-power of the shaft on the differential system, and $\scriptstyle \ \Phi$ that of the differential system on the governor; then the equation of motion becomes

(58)	$\Theta \delta \theta +\Phi \delta \phi +\left(\Xi -L{\frac {d^{2}\xi }{dt^{2}}}+M{\frac {d^{2}\eta }{dt^{2}}}\right)\delta \xi +\left(H-M{\frac {d^{2}\xi }{dt^{2}}}+N{\frac {d^{2}\eta }{dt^{2}}}\right)\delta \eta =0$

and if

(59)	${\begin{array}{c}\delta \xi =P\delta \theta +Q\delta \phi \\\delta \eta =R\delta \theta +S\delta \phi \\\end{array}}$

and if we put

(60)	${\begin{aligned}L'=LP^{2}+2MPR+NR2&\\M'=LPQ+M(PS+QR)+NRS&\\N'=LQ^{2}+2MQS+NS^{2}&\\\end{aligned}}$